Which of the following is a 1-dimensional topological manifold

The following question is from my Smooth manifolds assignment and I am confused regarding it. So, I thought of asking it here.

Which of the following subsets of $\mathbb{R}^2$ are 1-dimensional topological manifold:

(a){ $(x,y)\in \mathbb{R}^2$ : |x|=|y|}

(b){ $(x,y)\in \mathbb{R}^2: |x|=|y|, y\geq 0$}

(c){ $(x,y)\in \mathbb{R}^2: |x|=|y|, y> 0$}

(d){ $(x,y)\in \mathbb{R}^2: |x|=|y|, y\neq 0$}

Intuitively, it seems that all 4 maps are haursdorff and 2nd countable and every point $p\in X$ has a nbd. U which is homeomorphic to an open subset of $\mathbb{R}^n$.

Answer is that (a) is not a manifold but rest are.

Now, checking each part by definition seems very lenghty also , I have just done 10 lectures on the course till now.So, knowledge of results is basic. So, is their some short and elegent way to approach this problem.

They are all subsets of a known manifold (the plane), so Hausdorff and second countable come basically for free.

The problem with (a) is the origin. There the topological space does not really look like $\Bbb R^1$, the way it does everywhere else. There are four "branches" that meet in the same point, and that's not homeomorphic to the real number line.

All the other spaces either take away the origin (c and d) or take away two of the branches at the origin (b), in either case resolving the issue.